# Chapter 2 - Kinematics in One Dimension - Exercises and Problems - Page 62: 60

(a) There are 2 chances to jump on the bus; first at 3.0 seconds and then at 6 seconds. (b) The person can wait a maximum time of 0.25 seconds.

#### Work Step by Step

(a) The distance covered when running must be 9 meters more than the bus moves; $v~t=\frac{1}{2}at^2+9$ $(4.5~m/s)~t=\frac{1}{2}(1.0~m/s^2)~t^2+9.0~m$ $(1.0~m/s^2)~t^2-(9.0~m/s)~t+18.0~m=0$ $(t-3.0)(t-6.0)=0$ $t = 3.0~s, 6.0~s$ There are 2 chances to jump on the bus; first at 3.0 seconds and then at 6 seconds. (b) If we wait until the last possible moment, the bus's speed when we reach the door will be 4.5 m/s. We can find the time $t_b$ that the bus is moving. $(1.0~m/s^2)~t_b = 4.5~m/s$ $t_b = 4.5 ~s$ We can find the distance $x$ the bus moves in this time. $x = \frac{1}{2}at_b^2$ $x = \frac{1}{2}(1.0~m/s^2)(4.5~s)^2$ $x = 10.125~m$ The person needs to run a total distance of 19.125 meters. We can find the time $t_p$ it takes the person to run this distance. $t_p = \frac{19.125~m}{4.5~m/s}$ $t_p = 4.25~s$ The person can wait a maximum time of 4.5 s - 4.25 s which is 0.25 seconds.

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